Area and perimeter
🎯 In this topic you will
- Estimate and measure the perimeter of two-dimensional shapes
- Estimate and measure the area of two-dimensional shapes
- Determine the perimeter and area of composite shapes formed from two rectangles
🧠 Key Words
- area
- perimeter
Show Definitions
- area: The amount of surface a 2D shape covers, measured in square units.
- perimeter: The total distance around the boundary of a 2D shape.
Understanding Different Garden Shapes
T he shape of a garden is not always a rectangle. Knowing how to work out the perimeter of different shapes can help you to solve problems in everyday life.
❓ EXERCISES
1.
a. Measure the perimeter of the blue triangle.
b. Estimate the area of the blue triangle.
c. Measure the perimeter of the red triangle.
d. Estimate the area of the red triangle.
e. Copy and complete this sentence using the words ‘areas’ and ‘perimeters’.
The blue and red triangles have the same ____, but different ____.
f. Draw a rectangle with the same perimeter as the blue triangle.
g. Draw a rectangle with the same area as the blue triangle.
👀 Show answer
2.
a. Estimate the perimeter of this shape.
b. Measure the perimeter of the shape.
👀 Show answer
🧠 Think like a Mathematician
Both of these polygons have a perimeter of $16\text{ cm}$, but they have different areas.
Your Task:
- On 1 cm squared paper, use the grid lines to draw more polygons that have a perimeter of $16\text{ cm}$.
- Label each polygon with its area.
- Using whole numbers of centimetres on each side:
- What is the smallest area you can make with a perimeter of $16\text{ cm}$?
- What is the largest area you can make with a perimeter of $16\text{ cm}$?
- Check each shape carefully to ensure it meets the perimeter condition. This process is called specialising: drawing and testing specific examples to explore mathematical ideas.
Follow-up Questions:
Show Answers
- 1: Shapes can have the same perimeter but different arrangements of side lengths, which changes how much space they enclose. A long thin shape has less area than a compact shape with the same perimeter.
- 2: The smallest possible area with a perimeter of $16$ occurs when the shape is very elongated (for example, a $1\text{ cm} \times 7\text{ cm}$ rectangle), giving an area of $7\text{ cm}^2$. Irregular polygons can produce areas close to this value but not smaller using whole-centimetre sides.
- 3: The largest area for perimeter $16$ using whole-number side lengths is produced by shapes closest to a square. A $4\text{ cm} \times 4\text{ cm}$ square would have area $16\text{ cm}^2$, but its perimeter is $16$ only if each side is $4$. Thus the largest area you can make is $16\text{ cm}^2$.
❓ EXERCISES
3. Find the missing length on each shape.
👀 Show answer
a. Missing length is $40\text{ m}$.
b. Missing length is $10\text{ mm}$.
c. Missing length is $1\text{ km}$.
d. Missing length is $30\text{ m}$.
e. Missing length is $3\text{ cm}$.
4. Work out the perimeter of each shape in question $3$.
👀 Show answer
a. Perimeter $= 2 \times (40 + 20) = 120\text{ m}$.
b. Perimeter $= 13 + 5 + 7 + 5 + 6 + 10 = 46\text{ mm}$.
c. Perimeter $= 14 + 5 + 10 + 4 + 4 + 1 = 38\text{ km}$.
d. Perimeter $= 30 + 15 + 10 + 5 + 15 + 15 + 5 + 5 = 100\text{ m}$.
e. Perimeter $= 14 + 4 + 8 + 2 + 8 + 3 + 14 + 9 = 62\text{ cm}$.
5. Work out the perimeter of these shapes.
👀 Show answer
a. Perimeter $= 3 + 1 + 1 + 1 + 2 + 2 = 10\text{ cm}$.
b. Perimeter $= 4 + 2 + 3 + 1 + 1 + 3 = 14\text{ cm}$.
c. Perimeter $= 2 + 5 + 1 + 1 + 1 + 6 = 16\text{ cm}$.
d. Perimeter $= 10 + 6 + 5 + 4 + 5 + 10 = 40\text{ cm}$.
6. Which two green rectangles from $A$–$F$ can you use to make shapes $a$, $b$ and $c$?
👀 Show answer
a. Shape $a$ is made from rectangles $B$ and $C$.
b. Shape $b$ is made from rectangles $D$ and $F$.
c. Shape $c$ is made from rectangles $A$ and $E$.
7. Sketch these shapes on plain paper. Divide the shapes into two rectangles. Work out the area of each rectangle. What is the total area of each shape?
👀 Show answer
a. Divide into rectangles $5\text{ m} \times 5\text{ m}$ and $4\text{ m} \times 3\text{ m}$. Areas are $25\text{ m}^2$ and $12\text{ m}^2$, so total area is $25 + 12 = 37\text{ m}^2$.
b. Divide into rectangles $10\text{ m} \times 3\text{ m}$ and $4\text{ m} \times 4\text{ m}$. Areas are $30\text{ m}^2$ and $16\text{ m}^2$, so total area is $30 + 16 = 46\text{ m}^2$.
c. Divide into rectangles $8\text{ km} \times 4\text{ km}$ and $5\text{ km} \times 2\text{ km}$. Areas are $32\text{ km}^2$ and $10\text{ km}^2$, so total area is $32 + 10 = 42\text{ km}^2$.
8. A room has these floor measurements.
How much does it cost to cover the room using each of these carpets?
👀 Show answer
The room can be divided into rectangles $3\text{ m} \times 5\text{ m}$ and $3\text{ m} \times 3\text{ m}$, giving areas $15\text{ m}^2$ and $9\text{ m}^2$. Total area is $15 + 9 = 24\text{ m}^2$.
Cost for each carpet:
• Carpet at $\text{\$}18\text{ per m}^2$: area $24\text{ m}^2$ so cost $= 24 \times 18 = \text{\$}432$.
• Carpet at $\text{\$}16\text{ per m}^2$: cost $= 24 \times 16 = \text{\$}384$.
• Carpet at $\text{\$}13\text{ per m}^2$: cost $= 24 \times 13 = \text{\$}312$.
• Carpet at $\text{\$}12\text{ per m}^2$: cost $= 24 \times 12 = \text{\$}288$.
• Carpet at $\text{\$}15\text{ per m}^2$: cost $= 24 \times 15 = \text{\$}360$.
• Carpet at $\text{\$}21\text{ per m}^2$: cost $= 24 \times 21 = \text{\$}504$.